INTEGRAL FORMULAS ON PROJECTIVE SPACE AND THE RADON TRANSFORM

BO BERNDTSSON We construct a variant of Koppelman's formula for (0, q)-forms with values in a line bundle, 0 (l), on projective space. The formula is then applied to a study of a Radon transform for (0, q)-forms, introduced by Gindikin-Henkin-Polyakov . Our presentation follows along the basic lines of Henkin-Polyakov [3], with some simplifications . Introduction In two papers ([2] and [3]), Gindikin-Henkin, and Henkin-Polyakov have developed a theory of a Radon transform for differential forms on domains in projective space . The two basic results are that the Radon transform defines an injective map on the_ Dolbeault cohomology groups (Le . that the Radon transform of a a-closed differential form is zero if and only if the form is 8exact), and an inversion formula which, among other things, gives a description of the image of the map . As a matter of fact [2] and [3] give two different inversion formulas, of which the latter works in greater generality and at first looks quite different from the earlier one . The aim of this paper is to give proofs of those results using weighted integral formulas (cf .[1]) . This is in no way radically different from the methods in [2] and [3], but in our view it leads to some simplifications . Let us explain briefly what they consist in . _ The starting point a-closed form, f, as in [3] is a "Koppelman's formula" which represents a f =aK(f)+R(f) so that áK(f) is "th_ a-exact part"and R(f) is a remainder which vanishes if f is a-exact . Then one goes on to show that R(f) can be related to the Radon transform . Henkin and Polyakov obtain this descomposition by lifting the form f to (the sphere in) Cn+l , by means of the projectivn map 7 : On+ 1 --> Pn


Introduction
In two papers ( [2] and [3]), Gindikin-Henkin, and Henkin-Polyakov have developed a theory of a Radon transform for differential forms on domains in projective space .The two basic results are that the Radon transform defines an injective map on t h e _ Dolbeault cohomology groups (Le .that the Radon transform of a a-closed differential form is zero if and only if the form is 8exact), and an inversion formula which, among other things, gives a description of the image of the map .As a matter of fact [2] and [3] give two different inversion formulas, of which the latter works in greater generality and at first looks quite different from the earlier one .
The aim of this paper is to give proofs of those results using weighted integral formulas (cf .[1]) .This is in no way radically different from the methods in [2] and [3], but in our view it leads to some simplifications .Let us explain briefly what they consist in._ The starting point a-closed form, f, as in [3] is a "Koppelman's formula" which represents a f =aK(f)+R(f) so that áK(f) is " t h _ a-exact part"and R(f) is a remainder which vanishes if f is a-exact .Then one goes on to show that R(f) can be related to the Radon transform .Henkin and Polyakov obtain this descomposition by lifting the form f to (the sphere in) Cn+l , by means of the projectivn map 7 : On+ 1 --> Pn and then using a classical formula in e"+ 1 .Instead of this we will construct an explicit kernel directly on P" .What this amounts to is that we don't need to integrate over the fibers of the map 7r, which makes our formula simpler and more explicit .As it turns out the freedom to chose weight factor enables us to treát fórms with values in the standard line bundles, 0 (l), on P" .From the form of the kernel one can then read off almost directly the definition of the Radon transform.We can also give a rather simple proof of the inversion formula by letting the weight vary.Finally we also note that the different inversion formulas of [2] and [3] can be viewed as coming from different choices of weight .
Once again, it should be stressed that we have no results that are basically new, but that the difference lies mostly in the presentation .Since thus our aim is partly expository we have tried to make the paper as selfcontained as possible, starting with a preliminary section with background material about projective space .
As is the case in [2] and [3] our paper deals with forms of bidegree (0, q) or (n, q) .The construction of integral kernels on P" in the general case of (P, q) forms is more difficult .In a last section we sketch how a recent idea of C. Laurent-Thiebaut and J.P. Demailly can be used here.They showed how one can construct kernels on a complex manifold given a holomorphic connection on a complex vector bundle over the manifold .In their formalism, however, there appear certain "parasitary terms", which depend on the curvature of the connection .It turns out that, in the case of P", the curvature that enters has a particularity simple form, which enables us to push their construction one step further and get rid of the undesirable terms.We have not applied this construction to the problem of the Radon transform, but give it mostly as curiosa .

. Preliminaries on P"(O)
P" (C) (or briefly P") is the set of equivalente classes "+1 where if there. is a nonzero complex number A such that z = \w .
We say that z = (zo, ---z") are the homogenous coordinates of the point [z] .A hyperplane in P" is a set of the type " II e = {[z]; 1: Ijzi = 0} 0 where ~E Cn+1 \ {0} .Similary, a q-dimensional plane in Pn is an intersection of (n -q) hyperplanes in general position, Le. a set of the type n where (em) is linearily independent .m = l, . ..,nq), is a well defined function on Cn+l which is homogenous of degree l in z.Conversely the same equation defines a section to 0 (l) if f is l-hómogenous in z.
Hence we may identify sections to 0(l) with l-homogenous functions on Cn+ 1 .In particular a homogenous polynomiál of degrée l defines a holomorphic section to 0 (l) if l > 0.
Next we consider differential forms on (open sets of ) Pn .Since 7r : e n,, ---> pn '7r ( z ) = z is surjective, (or rather Since d7r is surjective) the pullback, 7r`, of forms on Pn to C n+ 1 is injective .Therefore we can identify a form on p n with its pullback, which simply means that we write the form in homogenous coordinates .If for instance we chose inhomogenous coordinates z1 zn zo zo : on Uo, a (1, 0)-form, on Uo can be written where the coefficients fi can be considered as 0-homogenous functions of z. for all forros g, and moreover the coefcients of f are homogenous of degree (-p, -q) in z and z respectively .
Here the brackets stand for the scalar product on forros induced by the standard metric on C n+ 1 (the forros dz, n dzj = dz1 n . . . .dzi, n . . . .dzi, form an orthonormal basis on (p, q) forms if (zo, . . ., zn ) are orthonormal coordinates on Cn+ 1 , and forros of different bidegree have scalar product zero) .
It is enough to verify this claim at a given point p E C"+ 1 , and we can choose orthonormal coordinates so that The operation of interior multiplication is defined by if f, h, g are forms.Thus, we see that if f is projective and h is an arbitrary (r, s) form whose coefficients are homogenous of degree (s, r), then hj f is again projective.
Let us finally study forms on Pn with values in 0(I) .Such a form can be thought of as a scalarvalued form fi on each Ui such that fi = ( zk)l fk zi in homogenous coordinates .So, written in homogenous coordinates a form of bidegree (1, 0) with values in 0 (l) is a form where a) Eo Fi zi = 0 and b) Fi are homogenous of degree l -1 in z .On Ui we can consider the (n, 0)-form This is clearly a nonvanishing (n, 0)-form and a simple computation shows that Hence Since any scalarvalued (n, 0)-form f can be written where the fi must satisfy k k dz0 wi = ~(-1) 0 i = (zk)n+'wk zi ff'wi fi = (zi )n+1 fk zk we see that the bundle of (n, 0)-forms (n, q)-form with values in 0 (l) can be in 0 (l') where (hj f, g) = (f, !A g) on Ui fl Uk, i. e. on Ui on Ui fl Uk is isomorphic to 0(-(n+l)) .Similary a identified with a (0, q)-form with values

Integral kernels on Pn
In this section we will construct integral kernels for the á-operator on domains in P" .They will operate on forms of bidegree (n, q) or (0, q) with values in a line bundle 0 (l), and satisfy a version of Koppelman's formula .In [3], Henkin and Polyakov have obtained such formulas by pulling forms on Pn back to the sphere in e n + l using 7r* .Here we will work directly on Pn writing our kernels in homogenous coordinates .First we recall the construction of weighted integral kernels in en (cf.[1]).
Consider a domain D C en with smooth boundary.Let s=(si~,sn) :DXD~e n be a Cl-function satisfying is(S,z)j C CiS -z1 and j(s,Sz)j ?elS-zlz uniformly for E D and z in any compact subset of D .Here for S, q E en (S,0 = L. S, ni . Then let Q :DxD-->~n be any C1 -function .With s and Q we identify the (1, 0)-forms which we denote by the same letters .' Let G(t) be an holomorphic function of one complex variable defined in a region such that the following formulas make sense, and which satisfies G(0) = 1 .We then define Here G( k) is the k-th derivative of G, and the a is taken with respect to 5 and z.Let K,,, be the component of K of bidegree (p, q) in z and (n -p, n -q -1) in S, and define Pp,q in the same way.We then have the following result (cf .Theorem 1 and Theorem 5 of [1]) : Here K,,,~, =: 0.

D D
To be quite accurate the theorem is stated in (1] only under.the assumption that Q(S, z) be holomorphic in z in which case P., q vanishes for q > 0 .However the same proof gives the result as formulated here.
We refer to (3) as Koppelmans formula and our first objective is to generalize the construction to domains in projective space.First we consider the case of (n, q)-forms .The construction follows precisely the same pattern as in the case of C" .Wé write the kernels in homogenous coordinates, using two d_ifferential forms s and Q .Tb.e novelty is that we must make sure that s and óQ define projective forms .Therefore we suppose as given a domain D in Pn and two functions S,Q :DXD-~C n-11 satisfying : i) s(S, z) is homogenous of degree -1 in S and 0 in z. ii) (s, z) = Eo s, zi = 0. iii) is(S, z) 1 < Cd(S, z) and 1 (s, S) ?cd2 (S, z) uniformly for S E D and z in any compact part of D.Here d is any distance function that behaves like the euclidean distance in local coordinates (a specific choice will be given later but is unimportant so far) .iv) Q(S, z) is homogenous of degree -1 in z and 0 in S.
As before we denote also by s and Q the forms s n and Q = E Q; dzj 0 (since we are dealing with (n, q)-forms we use dzj instead of d(Sj -zj )) .If conditions i)-iii) are fullfilled s is a projective (1, 0)-form with values in 0 (1) when considered as a form in z (cf.a) and b) of sec .0) .As a form in S it is of degree (0, 0) and takes values in 0 (-1) .Clearly Q is not a projective form but we claim that aQ is.It is enough to prove this in each 7r-1(Ui ), so assume e. g. that z0 7É 0 and put Q = (Qo -ZO ,Q1, ..,Qn)-

Then ¡he form
Considered as and as a form in now state: _ n Q = 1: Qi dzj 0 is a projective form in z with values in 0 (0), hence the which equals aQ .Note that aQ also takes values in 0 (0) In analogy with formulas ( 1) and ( 2) we put same holds for a Q, as a (0,0)-form in S .
a form in z the k-th term in K takes values in 0 (l) with z in 0(-l) .As it is easy to see the same goes for P. We can Theorem 1 .Let f be a (n, q) form in C1 (D) with values in 0(1) where l <_ n.Put N = n -l and define K9 and P9 as the componente of bidegree (n, q) in z and (0, n -q -1) in S of K and P respectively .Then f (S) n P q (S, z) )Í Proof. .The hypothesis quarantees that f n Kq etc are scalarvalued as forros in S so the integrals make sense.By a partition of unity we may assume that f has its support in one of the sets Uj , say in Uo .We may then also assume z lies in Uo since Uo is dense in P" .Put by i) .In the same way We also have and and K*, P* as so that Since S'/So and z'/zo are inhomogenous coordinates on Uo, (6) follows directly from Theorem 0, applied to the form Só -n f.M So far we have considered only (n, q) forms.The case of (0, q) forms is dual.One way to look at things would be to change the role of the variables .So let us define S * (S, z) = s(z, S) ) and P* (S, z) = P(z, S) .Then we find, either by repeating the proof of Theorem 1 or just by taking duals of (6), Theorem 1'.Let f be a (0, q) form in C1 (D) with values in 0 (l) where (6') f (z) = Cn,9 {la f(s) A Kq (s, z) -ID af (s) A K9 (s, z) Remark .Recall that the bundle of (n, 0)-forros on W" is isomorphic to 0 (-(n -}-1)), so that a 0 (l)-valued (n, q)-form can be considered as a 0 (l -(n -h 1))-valued (0, q)-form .Then we see that Theorem 1 gives a Koppelman formula for (0, q)-forros with values in 0(1) for l <_ -1.Hence, together with Theorem 1' we have formulas for all the line bundles 0(l), and actually both formulas work if -n < 1 < -1 .
We shall now exemplify Theorem 1 by giving a choice of s and Q that works for D = P" .As Q we can take We claim that To chose s we start by deciding what so that Q = a log IzI2 .
(s, S) = : -D (S, z) should be.By analogy with the Bochner-Martinelli formula in e" we take for 1 the square of the distance between S and z .
If S an z are in C"-1-1 we put where ~n z denotes the exterior product and we take norms in n a e n+ 1 .Thus oD(S, z) = 0 precisely when [S] = [z] .Then put (7) (s, S)=1 and (s, z)=0.
For this, note that for z fixed 1 is a quadratic form The corresponding bilinear form is the natural distance function on P" .Using those choices of s and Q we obtain the following well known corollary to Theoem 1 and 1' (cf Theorem 1 .4 in Corollary .For the Dolbeault cohomology groups of pn we have : In case 1 >_ 0 we have the following representation formula for an element of H0,0(pn, 0(1)) Proof.If f is a-closed a (n, q)-form and 1 < n, Theorem 1 gives , so a) follows by identifying f with a (0, q)-form with values in 0 (1 -(n + 1)) .In the same way b) follows from (6') since P* is of bidegree (0, 0) in z.Formula (8) follows from (6') when q = 0.

. q-concave domains in Pn
Let D be a domain with smooth boundary in Pn .We say (cf.[3]) that D is q--concave if for each point [z] E D there is a q-dimensional projective plane containing [z] that lies entirely within D. We assume moreover that the plane can be chosen so that it depends smoothly on We shall describe all the q-planes in D. Any such plane can be given as where A is (n -q) x (n + 1) matrix of maximal rank.Since 11 C D (1) Az = 0 -~, Iz I ) Iz11I .

Write A= (A', A")
where A' is of order (n -q) x (q + 1) and A" of order (n -q) x (n -q) .Then Az = A'z' + A" z" .
From this we see that A" must be nonsingular since otherwise some point z = (0, z") would lie in D contradicting (1) .Hence we may multiply A from the left by (A") -I which doesn't change the plane II, i. e. we may assume A" = I.Then (1) says that JIA' 1I < 1 .In conclusion we see that the set of q-planes in Dq can be identified with the set of (n -q) x (q + 1) matrices of norm strictly less than 1 .
Let [S] be a point in D9 and let As be the matrix corresponding to the linear map zr ---, -z o .S 1 " Is'12 5 .
Since S E Dq , IIAÍ 11 < 1. Moreover As S' + 0 so [S] lies in the plane defined by A, .Hence D9 is q-concave .
We shall now construct a representation formula for (n, r) forms in a qconcave domain .
By hypothesis we can find, in a smoth way, for each [z] E D a q-plane, n(z), such that At least locally, II can be represented at the set of S such that n (~k (z),S) _ j(z)Sj = 0, k = 1, . ..,(n -q), 0 where the functions {k are smooth and homogenous of degree 0. Then we can choose, locally in z, functions cpl , . . ., cp" _ q depending on ~and z such that but we shall see later that the choice is not important .Let us note however that we can also assume that cpk is homogenous of degree -1 in S and 0 in z.By means of a partition of unity in z we can finally find a function S' :Df xD.,C" +1 (3) (S' (S, z), S) > 0 with equality iff [S] E II (z) .
Moreover s' can be represented as i locally in z.Hence, for [S] E áD s' satisfies the requirements i) --iii) of the previous section .Our next objective is to modify s for [S] E D so that those conditions will be satisfied for all [S"] E D .
To this end we let s" denote the choice of s of the previous section i. e.
where p is a function that is positive inside D and vanishes to order 1 on aD .Now we can define the kernels K and P by formulas ( 4) and (5) of section (1) .We will leave the choice of Q open for the time being, requiring however that Q be independent of S (and homogenous of degree (-1) in z as always) .Recall that this is the case for the "standard choice" Q = aiog Izl z .
Let us now see what the representation formula in Theorem 1 gives when r <_ q (we are dealing with forms of bidegree (n, r) in a q-concave domain) .We also suppose q < n, since q = n means D = P", which is the case treated in section 1.
Since the kernel P is of pure bidegree (n, n) in z we have P,, = 0. Hence the only obstruction to solvability of the ó-equation comes from the boundary integral which we will now evaluate .In our definition of s, (5), the second term gives no contribution when S E aD .This is clear if p is not differentiated, but it also holds for the differentiated term since f is of full degree in ds so that 5p can be replaced by dp in the integral, and dp = 0 on aD .
Hence s can be replaced by s' in the calculation of and we will then use (4) .We find where we write Ck also for the form From this it follows immediately : We are looking for the component of K that is of degree (n -r -1) in ds and only the third term in (6) contributes .Hence the part of S' h (as, )n-k-1 that we are interésted in must contain a factor i :~Okek n (EÓSrk n SCk)n-r-1 í , which means that we have a product of (n-r) of the forms Stk, k = 1, . . ., (n-q).Thus, if r < q, some form {k must occur twice ; whence Kn ,, = 0 for S E áD if r < q.
Theorem 2. Let f be a a-closed form of bidegree (n, r) in the q-concave domain D,-taking values in the line bundle 0 (l), where l <_ n .Assume r < q.Then f is a-exact and we have f (z) = cn {áz f f(S) A K,,, r (S, z) } D See also [3,Theorem 2.2] .In other words if r < q, l <_ n.The first bidegree where we have non-vanishing cohomology is r = q which we shall study in the remainder of this paper.If r = q, (7) equals where and

Our kernel is
Hence we find for S E aD (10) = an,q k+Jal=q k<N Kn,q = bN,k,a (Q, S) N Hn,r (D, 4 (l)) = 0 an,g w q ( ;0) n S1 n . . . .en -q an,q = (n -q -1)i(-1)(n-q+1)(n-q)/2 n-q and we are interested in K n , q when S E aD .Let us use the notation =(,P1)a' . ..,(~Pn-q)a^-v lf a=(al, " ..,an-q) Since all the forms ¿m occur exactly once in (8), the first term in (6) will not give any contribution when we compute Theorem 3. (cf.(3, Theorem 2.21).Let f be a ó-closed form of bidegree (n, q) in the q-concave domain D, taking values in the liase bundle 0 (l), where l < n.Assume r < q .Then f is á-exact if and only if T(f) = 0 where In that case ProoL It is clear that if T(f) _ = 0 then f is á-exact so all we have to prove is that if f is ó-exact then T(f) = 0. Assume Then Hence f n K n,q = (-1)n+q g h asKn,q = (-1)n+, 9 n drKn,q aD aD aD But the kernels K and P satisfy and we are done.so that T(f) (z) = f D f(S) A Kn,q (S, z) .a f( Z ) = 5.Cn,q f f(s) n Kn,,-1(s, z) .Considering the components of bidegree (n, q) in z and (0, n -q -1) in S we get d,Kn,q+d .Kn,q-1 =0 if we recall Pn ,, = 0, r < n.But we have already noted that K,,,, = 0 for S E óD and r < q.
ds K ,q = 0 for S E áD If we put further restrictions on the homogenuity, l, the condition T (f) = 0 can be simplified a lot .We assume that 1<n-q N=n-l > q.
Actually an analogous discussion could be carried out for all N >_ 0, but the case N > q is simpler so we will be content with that case.
aD (p, e~S ) n-9 A few comments are in order .First, N is chosen so that the form we integrate takes values in 0 (0), so the integral makes sense .Second, if q = n -1 and l = 0 (so that N = n) where D* is the set of ¿ such that the q-plane IIF , defined by the equations (Em,S) = 0 m = 1, . ..,nq, lies in D. This quarantees that F(f)(e,rl) is well defined and holomorphic in ¿ if we make an appropiate choice of cp .In case q = n -1 it is evident that the definition of F(f) does not depend on the choice of (p.We shall now prove: Proposition 4 .The definition of F(f) does not depend on (p, provided óf = = 0. Proof. .Assume we have two choices (cpó ) and (~P-) m = 1, . . ., n -q.Thus (~p') are functions of ¿ and S such that (P9 ,1, S) 9~0 for 1 E D*, S E 9D, s = 0,1 .
For future reference we now note an invariance property of 3(f) .Consider e = (¿1 , . . ., ¿n-q) as a (n -q) x (n + 1) matrix, cp as a row matrix and ~as a colum.Let g be a nonsingular (n -q) x (n -q) matrix.Then ge defines the same q-plane as £, so let us see how 3(f) changes if we change e to g¿.Note that where Note that Thus (9) implies that The reason for the introduction of 3(f) was its relation to the operator T(f) .More precisely, the term a = 0 in the definition of T(f) could be written const.F(f)(¿,11)e1 A . . ..In-q A (di7)glf=f(zl .
Then f is ó-exact if and only if 3(f) = 0.
Proof.It follows from ( 16) and Theorem 3 that if ."{(f)= 0 then f is 8-exact .On the other hand if then 3(f) = 0 since by Lemma 5. f = ág a w, (v) = 0 «P, e, S)n -q Hence the Fantappié transform induces a one-to-one map on the cohomology groups Hn ,q (D, 0 (l)) if l < n -q .
For a á-closed form the Fantappié transform coincides with the Radon transform which we shall now define .
Let el, . ..en-q be a linearly independent set which define the q-plane II¿ .We shall now change the convention of the previous paragraph and identify ~m with the forms Definition 8. Let f be a (n, q) form in D with values in 0 (1), l < n -q .
Put N = n -l .Then the Radon transform of f is defined by

. The Radon transform n
Cm = 1: ¿m ds) .S 00 el A . . .¿ n-q J f(S) N-q IS1 n . ..en-q 12 (rl~S) Note that since f is of bidegree (n, q), we obtain a form of bidegree (q, q) after taking inner products with the form 11 n . . .en -q .Thus integration over the q-plane IIF is a well defined operation . .Theorem 9. Suppose f is a ó-closed form of bidegree (n, q) with values in 0 (1),1 < n -q.Then (2) Proof. .From the definition of R(f) it is clear that R(f) satisfies the same transformation law under a change of frame as does 3(f) .Namely, if g E E GL(n -q) R(f)(ge,rl) = (detg)-1R(f)(¿,rl) (cf.Prop .6) .Hence we may assume that el . . .¿"g is orthonormal, and we can chose orthonormal coordinates on C" +1 so that 11 = (0, . . .,1, 0 . . .0) are the last (n -q) elements of a dual basis .If we write S' = (So,---Sg ) S" = (Sg+1,-..Sn) the plane Iie is then given by S" = 0.In the definition of 3(f) we now take where dV is surface measure on H¿ .From this it follows that where f is the coefficient of dw 1 n . . .dwq n dw1 A . . .dwn of the form w0 a f .But this is precisely so we are done.
Prooi a) follows from Theorem 9, b) and c) are obvious from the definition, as is d) since (t¿, S) = 0 on II¿ if 1c lies in the span of t1 . . . .en-q .
Let io be a function that satisfies a) -d).Then ft,p takes values in 0(l) .Observe that the condition (e', z) = 0 guarantees that ft,r is projective, and that a section, t, to E is a map where still k+J«l-q k<N [z] ([Z],n(z),e(z» where g(z) is homogenous of degree -1 and e(z) is homogenous of degree 0. Note also that if g E GL(n -q) and we define by 9([z], 17, e) = ([z],rl,g¿), then 9* (np) = ft,p, by c) .In particular the pullback of f2,p under any section t, depends only on the plane IIF defined by e and not on the choice of frame.
Inspired by formula (16) of section 2 we consider together with ft,p the form It is not hard to check that ft,p also satisfies the invariance property g * A4) = ño if g is a matrix in GL(n -q) .As a matter of fact this even hólds if g depends on z, since each differentiation of g produces a factor e' and we already have a complete set of those forms.
From this it follows that if is a section to E the pullback t :D-E t* A+) only depends on the q-plane defined by j(z), and not on the choice of frame.
In particular, if (tj is a set of local sections to E such that the covectors es define the same q-planes on overlaps we have that is a global form.In the sequel we will not distinguish between two sections if they define the same q-planes. _ From the previous section it is clear that if 0 = R(f), where f is some 8-closed form, and if t is some section to E then Thus, in addition to the properties a) -d) of Proposition 10, ip has to satisfy a certain differential equation .
Proposition 11.Let ifi be a function satisfying properties a) -d) of Proposition 10.Then the following conditions are equivalent: el) If t is any section to E then at* (ñ p) = 0. e2 ) On E, dñp = 0. es ) zp satisfies the equations (In case 1 = n -q, ifi is independent of rt so the second set of equations is void.If moreover q = n -1 the first set of equations is also void.) It is clear that e2 implies el, since Moreover it will follow from the inversion formula that we shall now prove that el implies e3 , since it is obvious from the definition of .írthat any 0 that can be written will satisfy es .What remains to be proved is thus that we have found no slick proof of this fact we have resorted to the method of brute computation, some indications of which will be given after the inversion theorem .
In the proof of the inversion formula we assume that D satisfies one extra assumption.be two smooth functions that associates to S a q-plane II that lies in D and contains S. Assume Ih is defined in a neighbourhood of aD and II2 in a neighbourhood of a given q-plane Ro in D. Then there is a similar function SED such that II = Hl asear aD and II = II2 asear IIo .
e3 implies e2 .Since Clearly assumption (A) holds if the set of all planes passing through a given point is contractible, which is the assumption in [2] .(Hence in particular it holds for D = Dq see example at the beginning of section 2) .In [31 the inversion theorem is stated without extra assumption but no proof is given ._Theorem 12.For a function 0(e, rl) to be reprentable as 0 = R(f) with f ó-closed it is necessary and su icient that 0 satisfy properties a) -d) of Proposition 10 and property el of Proposition 11 .In that case if t is any section So let t be a section to E and fix a q-plane II( o) in D .Say II(o ) is given by the equations and as e = e(S) we will take 0 = c ,a R(f) with f = t' (ño ) t .D-->E.
Proof. .All that remains to prove is that if zp satisfies conditions a) -d), e l ) then y, = c ,9 R (f) with f = t-(ñ,4 ) .
By assumption el, óf =0, hence by theorem 9. Thus the value of R (f) at p depends only on the choice of t near aD, so by assumption (A) we can chose t any way we like near f(0) .As 17 = i7 (S) we take as usual e(S) = p when Fortunately enough, when ~E II,,, S, +m = 0 for m = 1, . . .n -q, so the only term in f that does not vanish on II,, is the one with ce = 0.When S E II,, and e(S) = p.Therefore and f = CN,q,010(¿, rl)(aa log Is'I2)q A dsq +l n . . .d~q+m, Nq,00(/U, I s 1 2)(aalogls'1 2 )q, . . .R(f) (p,6)-CN,q,0J 0(,u, Is l12 )(ó',s')N -q(áalogls'1 2 )q .fn f Recalling that 0 is a homogenous polynomial of degree n -lq = Nthe variable i? we find so the proof is complete.R(f)(u, 6 ) =Cn,gIP(P,a') (this is e.g. a consequence of (8) sec. 1) .But now property d) gives that vG(w,a') =VG(m,6) q in If ij j (S) _ 1 s ', the inversion formula we obtain is close to the one in [3] .But of course we are free to chose 71 in many other ways.If q = n -1 we can take This formula is the starting point of [2] .
Sketch of proof of Proposition 11 : As already remarked e2 implies el directly.Moreover since we only used property el in the proof of Theorem 12, we have that el implies that (~, rl) = c f A wQ (`P) (rl, S)N -q 8D (,p, e, S) n-q for some f .But then e3 follows directly by differentiátion under the integral sign .Thus all we need to prove is that e3 implies e2 .Here is an indication of one way to see this .Let  Inserting i), ii) and iii) in (6), and using the obtained relation in (5) we get dñ,p = 0. E

. Formulas for (p, q) forms
To start with, let M be a complex manifold' and let E be a holomorphic vector bundle over M. We suppose E is endowed with some hermitian metric and let v be the unique holomorphic connection in E compatible with the metric.Together with E we consider the dual bundle, E*, and denote the pairing between E* and E by The fact that v is holomorphic means that where v' is an operator of bidegree (1, 0) .Since v is the canonical connection of a hermitian metric the fact that a2 = 0 implies that Hence If, with respect to some holomorphic frame then so that v=a+v' OE 7, = (v9 + áv')17 .v'=a+0 a(v'71) = aa7, + (ae)i1-0 A áil = -v'(ái,) + (ae)91, OE q1 = (a0)71 .
Thus OE operates on a section as multiplication with a matrix of (1,1) forms.We now make the very restrictive assumption that this matrix is a multiple of the identity matrix, i .e. that OE yl = ri O, where 0 is a (scalarvalued) (1,1) form and multiplication is componentwise.n f (s) A P,,, (S, z) .
Since l = 0, N = p.This implies that in the definition of P the terms with /i, > p vanish .Moreover the terms with Pi < p cannot have degree p in dz.llence Pp = c(aQ, dz)p A E)" -p = = c(aalog Izlz)p A (aalog Isi 2 ) ^-p which immediately gives the statement .
Pn is covered by the (n + 1) open sets Given such a section Ui = {[z]; zi ,~0}, each of which is biholomorphic to C n under the map' zo zi-1 zi+ 1 zn zi zi zi zi The complement of Ui in pn is the hyperplane {[z] ; zi = 0}, sometimes called the hyperplane at infinity (relative to Ui ).For each integer l we have a line bundle, 0(l), over pn defined by the tranto 0 (l) is a function f i on each Ui such that f 7 =gikfk on Uinu k .
Fi zi = 0 and b) Fi are homogenous of degree -1 in z.If, conversely, Fi satisfies a) and b) the lame relation can be used to define a form on U o and a similar relation works on any Ui .Hence a) and b) characterize the projective forms of bidegree (o, i), i. e. the forros on e n, 1 in the image of ir* .More generally we have:A form f on C n + 1 of bidegree (p, q) is projective if and only if [p] = [1, 0 . . . .0] .We take zi z o as inhomogenous coordinates on Pn near [p] .At p dwi = dzi .zo Hence a projective form will not contain dzo or dzo which means that (1) holds.If conversely f does not contain dzo , dzo f = j:fjjdzl Adzj = j :(frjzózo 9)dwI Adwi at p which defines f as a form at [p] .
nq ~(p k (S, z) el (z), S) ?0 with equality if and only if [S] E 1I(z).One possible choice is such that and let (pk = W, O/IS1 2 IP) A (je)0' A (5Q)k A S1 A . . ., S n q (~P, e, S) n-k where a,,,, is some constant (not the same as before) and bN, what makes Koppelmans formula true) .